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Stoichiometry in solution
Combining n = cV with mole ratios to find masses of precipitates, volumes of reacting solutions, limiting reagents and leftover ion concentrations in solution, a concentration from a precipitate mass, and the method of continuous variation for finding a mole ratio by experiment.
Most reactions in a lab, in your body and in the ocean happen between dissolved substances. To predict how much product forms, you can't weigh the reactants directly: they are spread through a liquid. But if you know the concentration and the volume, you know the moles, and the rest of the calculation is the same as in 2D.
The method
Stoichiometry in solution adds one more entry point to the mole map: for a solution, n = cV.
- Find moles of the known substance: n = cV for a solution, n = m ÷ M for a solid.
- Use the mole ratio from the balanced (or net ionic) equation.
- Convert to what is asked: m = nM, c = n ÷ V or V = n ÷ c.
50.0 mL of 0.200 mol L⁻¹ barium chloride is mixed with excess sodium sulfate solution. What mass of barium sulfate precipitates?
Net ionic equation: Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s).
n(Ba²⁺) = 0.200 × 0.0500 = 0.0100 mol; ratio 1 : 1, so n(BaSO₄) = 0.0100 mol.
M(BaSO₄) = 137.3 + 32.07 + 4 × 16.00 = 233.4 g mol⁻¹; m = 0.0100 × 233.4 = 2.33 g.
What volume of 0.250 mol L⁻¹ hydrochloric acid reacts exactly with 2.00 g of calcium carbonate? CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g)
n(CaCO₃) = 2.00 ÷ 100.09 = 0.01998 mol; n(HCl) = 2 × 0.01998 = 0.03996 mol.
$V = \dfrac{n}{c} = \dfrac{0.03996}{0.250} = 0.160\ \text{L} = 160\ \text{mL}$
Limiting reagents in solution
When two solutions are mixed, find the moles of each reactant from n = cV and compare (moles ÷ coefficient), just as in 2D. After the reaction, the excess ions are spread through the combined volume.
25.0 mL of 0.400 mol L⁻¹ AgNO₃ is mixed with 30.0 mL of 0.250 mol L⁻¹ NaCl. Calculate the mass of AgCl formed and the concentration of silver ions left in solution.
Ag⁺(aq) + Cl⁻(aq) → AgCl(s). n(Ag⁺) = 0.400 × 0.0250 = 0.0100 mol; n(Cl⁻) = 0.250 × 0.0300 = 0.00750 mol. Ratio 1 : 1, so Cl⁻ is limiting.
n(AgCl) = 0.00750 mol; M(AgCl) = 107.9 + 35.45 = 143.4 g mol⁻¹; m = 0.00750 × 143.4 = 1.08 g.
Ag⁺ left = 0.0100 − 0.00750 = 0.00250 mol in 25.0 + 30.0 = 55.0 mL. [Ag⁺] = 0.00250 ÷ 0.0550 = 0.0455 mol L⁻¹.
Finding a concentration from a precipitate
The reverse problem is how chemists measure unknown concentrations: precipitate the ion completely with an excess of another ion, then filter, dry and weigh the precipitate. This is gravimetric analysis, which you'll study in detail in Year 12 Module 8.
Excess barium chloride is added to 25.00 mL of a sulfate solution. The dried BaSO₄ has a mass of 0.583 g. What is [SO₄²⁻]?
n(BaSO₄) = 0.583 ÷ 233.4 = 2.498 × 10⁻³ mol = n(SO₄²⁻).
c = 2.498 × 10⁻³ ÷ 0.02500 = 0.0999 mol L⁻¹.
Finding a mole ratio by experiment: continuous variation
How do chemists know that lead(II) and iodide ions combine 1 : 2? One method is continuous variation. Take two solutions of the same concentration and mix them in different proportions, keeping the total volume constant. Because the concentrations are equal, the ratio of volumes equals the ratio of moles. The mixture that gives the most product is the one where neither reactant is in excess, so it reveals the mole ratio of the reaction.
On the left of the peak, the first reactant is limiting, so the amount of precipitate rises in proportion to it. On the right, the second reactant is limiting. Drawing the two best-fit lines and finding where they intersect gives the ratio more reliably than reading the highest single point.
- Choose Mystery pair 1, predict which tube gives the tallest precipitate, then press Mix all tubes and let settle. The peak is at 4 mL of A : 8 mL of B. What formula does that suggest for the precipitate?
- Turn on Draw the two lines to see which reactant limits each side.
- Try pairs 2, 3 and 4. You should find ratios of 1 : 1, 2 : 1 and 1 : 3. Write a net ionic equation for each.
- Why must both solutions have the same concentration for volume ratio to equal mole ratio?
Test yourself
What mass of AgCl (M = 143.4 g mol⁻¹) forms when 20.0 mL of 0.150 mol L⁻¹ NaCl reacts with excess AgNO₃?
n = 0.150 × 0.0200 = 3.00 × 10⁻³ mol × 143.4 = 0.430 g.
What volume of 0.100 mol L⁻¹ NaOH reacts exactly with 15.0 mL of 0.200 mol L⁻¹ H₂SO₄? H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
n(H₂SO₄) = 3.00 × 10⁻³; n(NaOH) = 6.00 × 10⁻³ mol; V = 6.00 × 10⁻³ ÷ 0.100 = 0.0600 L.
10.0 mL of 0.50 mol L⁻¹ Pb(NO₃)₂ is mixed with 10.0 mL of 0.50 mol L⁻¹ KI. Which ion is limiting?
Both are 5.0 × 10⁻³ mol, but Pb²⁺ needs 2I⁻, so I⁻ runs out first.
In a continuous variation experiment with 0.10 mol L⁻¹ solutions of X and Y, the most precipitate forms with 9 mL of X and 3 mL of Y. The mole ratio X : Y is
Equal concentrations, so the volume ratio 9 : 3 = 3 : 1 is the mole ratio.
Put the steps in order for finding the mass of precipitate from two solutions.
For a solution, moles = concentration × in litres. After two solutions are mixed, leftover ions are spread through the volume. In continuous variation the two solutions must have equal , so that the volume ratio equals the ratio.
A student mixed different volumes of 0.200 mol L⁻¹ silver nitrate (X) with 0.200 mol L⁻¹ sodium phosphate (Y), keeping the total volume at 16.0 mL, and weighed the dried yellow precipitate.
| mL of X | 4.0 | 8.0 | 10.0 | 12.0 | 14.0 |
|---|---|---|---|---|---|
| mass of precipitate (g) | 0.112 | 0.224 | 0.279 | 0.335 | 0.168 |
(a) Analyse the data to determine the mole ratio of Ag⁺ to PO₄³⁻ and write the net ionic equation. (b) Use the 12.0 mL result to confirm the molar mass of the precipitate.
(a) The mass rises in proportion to the volume of X up to 12.0 mL (0.112, 0.224, 0.279, 0.335 g: about 0.028 g per mL), where Ag⁺ is limiting, and then falls at 14.0 mL because PO₄³⁻ is now limiting. The maximum is at 12.0 mL X : 4.0 mL Y. Both solutions are 0.200 mol L⁻¹, so the mole ratio Ag⁺ : PO₄³⁻ = 12.0 : 4.0 = 3 : 1. Net ionic equation: 3Ag⁺(aq) + PO₄³⁻(aq) → Ag₃PO₄(s).
(b) At 12.0 mL both are used up exactly: n(PO₄³⁻) = 0.200 × 0.00400 = 8.00 × 10⁻⁴ mol = n(Ag₃PO₄). M = 0.335 ÷ 8.00 × 10⁻⁴ = 419 g mol⁻¹. For Ag₃PO₄: 3 × 107.9 + 30.97 + 4 × 16.00 = 418.7 g mol⁻¹, which agrees.